# If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth terms

Given: t9 = 0

t9 = a + (9 – 1)d

0 = a + 8d

a = – 8d

To Show: t29 = 2× t19

Now,

t29 = a + (29 – 1)d

t29 = a + 28d

t29 = – 8d + 28d = 20 d (since, a = – 8d )

t29 = 20 d

t29 = 2 × 10 d ….(1)

Also,

t19 = a + (19 – 1)d

t19 = a + 18d

t19 = – 8d + 18d = 10 d (since, a = – 8d )

t19 = 10 d …..(2)

From eq. (1) and eq. (2) we get,

t29 = 2× t19

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